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Neglecting air resistance, the components of acceleration of an object that's dropped, tossed,

pitched, flung, lobbed, heaved, launched, or shot are constant.

The horizontal component is zero.

The vertical component is 9.8 meters per second2, directed downward.

These are both constant throughout the object's trajectory.

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Q: What are the horizontal and vertical components of the acceleration vector at the maximum height?
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When projectile has maximum horizontal range?

A projectile has maximum horizontal range when it is launched at an angle of 45 degrees to the horizontal. This angle allows for the ideal balance between the horizontal and vertical components of the projectile's velocity, ensuring that it travels the farthest distance before hitting the ground.


How does angle of projection affect the maximum height?

The angle of projection affects the maximum height by determining the vertical and horizontal components of the initial velocity. At 90 degrees (vertical), all the initial velocity is vertical which results in maximum height. As the angle decreases from 90 degrees, the vertical component decreases, leading to a lower maximum height.


In projectile motion What Angle would horizontal and vertical distance be equal?

In projectile motion, the horizontal distance will be at its maximum when the angle is 45 degrees. At this angle, the vertical and horizontal components of the initial velocity are equal, resulting in maximum range.


Is the maximum number of horizontal and vertical pixels that can be displayed on a screens?

That varies depending on the monitor. You will have to read the specifications of the specific monitor that you want the answer for. There is no fixed maximum value.


What are the two types of motion a projectile has?

If you throw ball at an angle above horizontal, you will see the path of the ball looks like an inverted parabola. This is result of the fact that the ball's initial velocity has a horizontal and vertical component. If we neglect the effect of air resistance, the horizontal component is constant. But the vertical component is always decreasing at the rate of 9.8 m/s each second. To illustrate this, let the initial velocity be 49 m/s and the initial angle be 30˚. Horizontal component = 49 * cos 30, Vertical = 49 * sin 30 = 24.5 m/s As the ball rises from the ground to its maximum height, its vertical velocity decreases from 24.5 m/s to 0 m/s. As the ball falls from its maximum height to the ground, its vertical velocity decreases from 0 m/s to -24.5 m/s. Since the distance it rises is equal to the distance it falls, the time that it is rising is equal to the time it is falling. This means the total time is equal to twice the time it is falling. This is the reason that the shape of the ball's path is an inverted parabola. At the maximum height, the ball is moving horizontally. If you do a web search for projectile motion, you will see graphs illustrating this.


How do you find projectile motion?

To find the projectile motion of an object, you can use the equations of motion for horizontal and vertical components. The horizontal motion is constant, while the vertical motion is affected by gravity. By determining the initial velocity, angle of projection, and acceleration due to gravity, you can calculate various parameters like range, maximum height, and time of flight.


A ball is thrown from the ground into the air at a height of 9.1 m the velocity is observed to be v7.6i 6.1j in meters per second x axis horizontal y axis vertical and up a to what maximum h?

To find the maximum height, we first need to separate the initial velocity into its x and y components. Since the initial velocity is given as v = 7.6i + 6.1j, the initial vertical velocity is 6.1 m/s. We can use the kinematic equation for vertical motion: v_f^2 = v_i^2 + 2aΔy, where v_f = 0 at the maximum height. Rearranging the equation to solve for the maximum height, h, we have h = (v_i^2)/2g, where g is the acceleration due to gravity (9.81 m/s^2). Plugging in the values, we find h ≈ 1.88 m.


How can you cut a piece of log to get maximum number of pieces from 3 cuts?

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At what angle from a horizontal should a weapon be fired to achieve minimum distance traveled by the projectile?

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How do you find maximum height for a projectile without knowing the time?

Assuming the simple model where the object is projected with an initial velocity of u metres/second at an angle of x to the horizontal, and that the only force acting on it after that is gravitational acceleration, g = 9.81 metres/second^2, then h = [u*sin(x)]^2/(2*g) metres.If the launch is vertical then x = pi/2 radians and h = u^2/(2*g) metres.


Why does a 45 degree throw go the furthest?

A 45-degree throw maximizes the horizontal distance traveled by balancing the vertical and horizontal components of the projectile's velocity. At this angle, the horizontal component is at its maximum, maximizing the range the object can travel before hitting the ground. Any angle higher or lower will result in a shorter distance traveled.


What is maximum acceleration rate?

the acceleration is equal to energy that release by the friction that came be electic that travel form somewhere.It proves that maximum acceleration rate.The easy explainationof that is Energy and Velocity are equal to maximum of acceleration